For what values of k will the equation have no real roots?

[Jaco Viljoen]

Homework Statement

2x^2-3x+kx=-1/2

1. k<1 or k>1
2. 1<=k<=5
3. k<=1 or k>=5
1<k<5

Homework Equations

b^2-4ac
a=2 b=3 c=k

The Attempt at a Solution

(3)^2-4(2)(k)
=9-8k<0
=9/8<k
=1&1/8<k

I get the answer above but don't know how it relates?
Any insight would be appreciated.

Thank you,
Jaco

 

[MarcusAgrippa]

Find the discriminant of your quadratic equation. The equation has no real roots if the discriminant is less than zero.

 

[jbriggs444]

Homework Statement

2x^2-3x+kx=-1/2

Homework Equations

b^2-4ac
a=2 b=3 c=k

Look at the original equation again.

What is the coefficient on x? What is the constant term?

 

[MarcusAgrippa]

Are you sure about your value for b?

 

[Jaco Viljoen]

Jbriggs,
I have been looking at other threads and found a similar example and there:
(-3)^2-4(2)(k+1/2)
=-9-8k+4
=-5-8k
=> k=5/8
Thank you for pointing that out Marcus

 

[Jaco Viljoen]

Is this correct?

 

[MarcusAgrippa]

It doesn't look correct. What is the general form of a quadratic equation? Write your equation in a way that imitates that general form and identify correctly the values of a,b,c. Then write down the general definition of the discriminant, and substitute your values of a,b,c into it. The solve the inequality (discriminant) < 0 for k.

 

[SammyS]

Homework Statement

2x^2-3x+kx=-1/2

1. k<1 or k>1
2. 1<=k<=5
3. k<=1 or k>=5
1<k<5

Homework Equations

b^2-4ac
a=2 b=3 c=k

The Attempt at a Solution

(3)^2-4(2)(k)
=9-8k<0
=9/8<k
=1&1/8<k

I get the answer above but don't know how it relates?
Any insight would be appreciated.

Thank you,
Jaco

Please restate the entire problem as it was given to you.

You may have a typo in the quadratic expression as you posted it. If it's correct as it is, then you still do not have the correct b or c .

 

[Jaco Viljoen]

Sammy,
For what values of k will the equation 2x^2 -3x + kx = -1/2 have no real roots?

Possible answers:
1. k<1 or k>1
2. 1<=k<=5
3. k<=1 or k>=5
1<k<5

 

[jbriggs444]

Everyone responding to this thread is attempting to point out that the b and c that you have harvested from that equation are wrong.

 

[Jaco Viljoen]

ok, I have been considering this too as:
(3x+kx)^2-4(2)(-1/2)
(3x+kx)(3x+kx)+4
9x^2+3kx^2+3kx^2+k^2x^2
9x^2+6kx^2+k^2x^2

 

[jbriggs444]

Re-read post #7 above. What is the standard form for a quadratic equation? Can you restate the original equation in that form?

 

[Jaco Viljoen]

ax^2+bx+c=0
2x^2+(3x+kx)+1/2=0
2x^2+3x+kx=-1/2
kx=-1/2-2x^2-3x
k=(1/2-2x^2-3x)/x

 

[Jaco Viljoen]

wow, i feel more confused...

 

[SammyS]

ax^2+bx+c=0
2x^2+(3x+kx)+1/2=0

It's good to this point.

Take the expression in parentheses and factor out x.

 

[Jaco Viljoen]

2x^2+(3x+kx)+1/2=0
2x^2+x(3+k)+1/2=0
2x^2+(3+k)(x+1/2)=0
2x^2+3x+1&1/2+kx+1/2k=0
kx+1/2k=-2x^2-3x-1&1/2
3/2kx=-2x+3+1&1/2
3kx=-4x+9
k=(-4x+9)/x

 

[SammyS]

2x^2+(3x+kx)+1/2=0
2x^2+x(3+k)+1/2=0

Stop at this point !

2x^2+(3+k)(x+1/2)=0
2x^2+3x+1&1/2+kx+1/2k=0
kx+1/2k=-2x^2-3x-1&1/2
1&1/2k=(-2x+3+1&1/2)/x

What is the coefficient of x ?

 

[Jaco Viljoen]

1?

 

[jbriggs444]

1

In the expression 2x² + 3x + 4, what is the coefficient on the "x" term?

 

[Jaco Viljoen]

3

 

[jbriggs444]

Good.

Now, in the expression 2x² + x(3+k) + 1/2, what is the coefficient on the x term?

 

[Jaco Viljoen]

2x^2+1x(3+k)+1/2

1 or is it still 3?

2x^2+1x(3+k)+1/2

 

[Jaco Viljoen]

I am missing this, I just can't seem to get it...

 

[Jaco Viljoen]

Its not the 2 is it?

 

[SammyS]

What if we write it like this?

2x² + (3+k)x + (1/2) = 0

What is the coefficient of x² ?

What is the coefficient of x ?

What is the constant term ?

 

[Jaco Viljoen]

2
3+k
1/2

 

[SammyS]

2
3+k
1/2

Now proceed.

 

[Jaco Viljoen]

x^2+(3+k)+1/2=0
like this?

 

[Jaco Viljoen]

x^2+(3+k)+1/2=0
a b c
x = (-(3+k) +/-√((3+k)2 - 4x^2(1/2)))/2x^2

 

[jbriggs444]

x^2+(3+k)+1/2=0
a b c
x = (-(3+k) +/-√((3+k)2 - 4x^2(1/2)))/2x^2

In the discriminant you where you are trying to write b^2 - 4ac. What was the value you came up with for a again?

 

[Jaco Viljoen]

(x+ )(x+ )=0

 

[jbriggs444]

(x+ )(x+ )=0

No. That is not correct and is not what SammyS agreed was correct. Refer back to your post #26.

 

[Jaco Viljoen]

jbriggs,
I don't understand what you are asking?

 

[Jaco Viljoen]

2
3+k
1/2

 

[jbriggs444]

2
3+k
1/2

That's a = 2, b = 3+k and c=1/2

Given that, what is the value of a?

 

[Jaco Viljoen]

x^2

 

[jbriggs444]

x^2

If a = 2, what is the value of a?

 

[Jaco Viljoen]

2

 

[Jaco Viljoen]

ok, just looking where this went wrong

 

[jbriggs444]

Right. a is 2.

So if you are going to substitute values into "b^2 - 4ac" you should put "2" in for a. You should not put in x^2.

 

[Jaco Viljoen]

2
3+k
1/2

(3+k)^2-4(2)(1/2)
=(3+k)(3+k)-4
=9+3k+3k+k^2-4

 

[Jaco Viljoen]

k^2+6k+5

 

[Jaco Viljoen]

(k+1)(k+5)

 

[Jaco Viljoen]

k=-1 and k=-5

 

[Jaco Viljoen]

1. k<1 or k>1
2. 1<=k<=5
3. k<=1 or k>=5
4. 1<k<5
5. none

 

[jbriggs444]

k=-1 and k=-5

Refer back to post #7.

You want values of k for which the discriminant is non-negative. You have identified values of k for which the discriminant is zero.

Edit: To be clear, you are on the right track and doing well now.
Edit: And reading back to the original problem statement we want values of k for which the discimant is negative.

 

[Jaco Viljoen]

It doesn't look correct. What is the general form of a quadratic equation? Write your equation in a way that imitates that general form and identify correctly the values of a,b,c. Then write down the general definition of the discriminant, and substitute your values of a,b,c into it. The solve the inequality (discriminant) < 0 for k.

 

[Mark44]

2
3+k
1/2

(3+k)^2-4(2)(1/2)
=(3+k)(3+k)-4
=9+3k+3k+k^2-4

k^2+6k+5

(k+1)(k+5)

k=-1 and k=-5

Jaco, instead of adding new posts, use the Edit button to made changes to an existing post.

 

[Jaco Viljoen]

ok Mark,

jbriggs,
I am not following you...?

 

[Jaco Viljoen]

2
3+k
1/2

(3+k)^2-4(2)(1/2)
=(3+k)(3+k)-4
=9+3k+3k+k^2-4
=k^2+6k+5

b^2 - 4ac

(6k)^2-4(k^2)(5)
(6k)(6k)-20k^2
36k^2-20k^2
16k^2